Thread: Points on a graph (tangent line finding)

1. Points on a graph (tangent line finding)

How can I determine the point(s), if any, at which the graph of the function has a horizontal tangent line?

Given: $y=x^2+2x$

Any help is welcomed

2. Originally Posted by hemi
How can I determine the point(s), if any, at which the graph of the function has a horizontal tangent line?

Given: $y=x^2+2x$

Any help is welcomed
hi hemi
you will get dy/dx. it is the slope of the tangent.
since in the question it has asked for horizontal line,means tangent parellel to x axis.
therefore, its slope will be zero.
put dy/dx=0
and get the value of x
put this value in the function and get the value of y.

thank you

3. Originally Posted by ursa
hi hemi
you will get dy/dx. it is the slope of the tangent.
since in the question it has asked for horizontal line,means tangent parellel to x axis.
therefore, its slope will be zero.
put dy/dx=0
and get the value of x
put this value in the function and get the value of y.

thank you
Hi,

Thanks for your post -- I'm still a bit confused though, the majority of this stuff is new to me - can you show me how I'd go about doing what you said so I can see and try to learn by examples? (This is how I learn best )

Thanks!

4. Originally Posted by hemi
Hi,

Thanks for your post -- I'm still a bit confused though, the majority of this stuff is new to me - can you show me how I'd go about doing what you said so I can see and try to learn by examples? (This is how I learn best )

Thanks!
hi hemi
y=2*x+x^2
dy/dx=2+2*x. this is the slope of the tangent.
since tangent is parellel to x-axis therefore
dy/dx=0
=> 2+2*x=0
=> x=-1
put this value in function to get the value of y
=> y=-1
therefore, the point on the function at which tangent of the function will be horizontal is (-1,-1).

Thank You

5. You could have done this, by the way, without using Calculus. This is a quadratic function so its graph is a vertical parabola. A tangent line will be horizontal only at its vertex. And we can get the vertex by completing the square: $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+1)^2- 1$. For x= -1 we have $y= 0^2+ 1= 1$. For any other value of x, x+ 1 is NOT 0 so $(x+ 1)^2$ is positive and y is greater than 1. (-1, 1) is the vertex.